Another rather simple Differential Equation

1. Jun 1, 2006

JoshHolloway

$$y \prime = (cos^{2}x)(cos^{2}(2y))$$

$$\frac{dy}{dx} = \frac{1}{2} (cox(2x) + 1) \frac{1}{4}(cos(4y) + 1)$$

$$\frac{1}{cos(4y) + 1}dy \ = \ \frac{1}{8} (cos(2x) + 1)dx$$

$$\int \frac{1}{ \sqrt{cos(4y)}^{2} + 1}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx$$

$$tan^{-1}(cos4y) \ = \ \frac{1}{8} \frac{1}{2} sin(2x) + \frac{1}{8} x + C$$

$$tan[tan^{-1}(cos4y) \ = \ \frac{1}{16} sin(2x) + \frac{1}{8} x + C]$$

$$cos^{-1}[cos(4y) \ = \ tan( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

$$y \ = \ \frac{1}{4} cos^{-1}[tan( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

Last edited: Jun 1, 2006
2. Jun 1, 2006

Hootenanny

Staff Emeritus
What's the question? What have you attempted?

~H

3. Jun 1, 2006

JoshHolloway

The first line is the problem, and each following line is one of the steps I took to solve it. The final line is the solution. Is that right? It sure doesn't look right to me, but I don't know what I am doing wrong.

Last edited: Jun 1, 2006
4. Jun 1, 2006

Hootenanny

Staff Emeritus
I am a little concerned as to what you have done with respect to this line here;

$$\int \frac{1}{{\color{red} \sqrt{cos(4y)}^{2} + 1}}dy \ = \ \frac{1}{8} \int (cos(2x) + 1)dx$$

Why did you change it into this form? Why did you simply not integrate;

$$\int \frac{1}{cos(4y) + 1}dy$$

In any case

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

~H

5. Jun 1, 2006

JoshHolloway

Why is this so:

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

And I changed it into that form because I thought that would be an easy way to use the arctangent integration technique.

How would I integrate this:
$$\int \frac{1}{cos(4y) + 1}dy$$

Last edited: Jun 1, 2006
6. Jun 1, 2006

JoshHolloway

Oh, would this work:

$$\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})$$

I reworked the problem that way, and this is what I ended up with:

$$y \ = \ \frac{1}{4} cos^{-1}[tan^{2}( \frac{sin2x}{16} + \frac{x}{8} + C)]$$

Am I getting warmer?

Last edited: Jun 1, 2006
7. Jun 1, 2006

JoshHolloway

I don't understand why this is so:

$$\frac{1}{cos(4y) + 1} \neq \frac{1}{ \sqrt{cos(4y)}^{2} + 1}$$

Isn't this correct?: $$cos(4y) \ = \ \sqrt{cos(4y)}^{2}$$

8. Jun 1, 2006

Hootenanny

Staff Emeritus
Sorry I didn't see you had raised it to the power two, my eyes must be getting worse!

I think the easiest way to do this is write it in terms of sec2.

~H

9. Jun 1, 2006

JoshHolloway

Does this work:
$$\int \frac{1}{cos(4y) + 1}dy \ = \ tan^{-1}( \sqrt{cos4y})$$

How would you write that in terms of $$sec^{2}$$

10. Jun 1, 2006

JoshHolloway

Wouldn't you have to use the method of Partial Fractions in order to get it into some form like this in order to write it in terms of sec
$$\frac{A}{cos4y} + 1$$

11. Jun 1, 2006

JoshHolloway

I have absolutely no idea what you mean by the easiest way would be to get it in terms of sec^2

12. Jun 1, 2006

TD

I'm having the feeling you're making this harder than necessary.
Remember that (tan(2y))' = 2/(cos²(2y)).

$$\frac{{dy}}{{dx}} = \left( {\cos ^2 x} \right)\left( {\cos ^2 \left( {2y} \right)} \right) \Leftrightarrow \frac{{dy}}{{\cos ^2 \left( {2y} \right)}} = \cos ^2 xdx \Leftrightarrow \frac{1}{2}\tan \left( {2y} \right) = \frac{1}{4}\left( {\sin \left( {2x} \right) + 2x} \right) + C$$

13. Jun 1, 2006

Hootenanny

Staff Emeritus
Yes, using partial fractions;

$$\frac{1}{\cos(4y) +1} = \frac{1}{2}\sec^{2}(2y)$$

Now, you sould find;

$$\int \frac{1}{2}\sec^{2}(2y) \; dy$$

A little easier to solve.

~H

14. Jun 1, 2006

TD

But we started with sec²(2y), that was the factor cos²(2y) at the rhs...

Perhaps I'm missing something but I think you just went in a circle, from cos²(2y) to the double angle (losing the square) and now back introducing the square, with the half angle!

15. Jun 1, 2006

JoshHolloway

Wow. Thanks ALOT! I really do have a tendancy to make things more difficult than they really are. Hey TD, could you check out the last question in the thread I started yesterday. Thanks.

16. Jun 1, 2006

Hootenanny

Staff Emeritus
Good point, guess my maths is a bit rough, best stick to physics...

~H

17. Jun 1, 2006

TD

We're all here to help (or to get help), I'd probably make similar mistakes when attempting to answer physics questions

It's nice to see you're helping arround a lot, being a science advisor and homework helper. I remember you from a couple of months back, with a few calculus questions

18. Jun 1, 2006

Hootenanny

Staff Emeritus
Hmm, so do I, volumes of revolution and integration by substitution (one where I actually forgot to integrate) once again, my careless mistakes let me down . Thanks again for helping me

~H